integral from 3/2 to 6 of x/(sqrt(2x-3))

解

\int_{\frac{3}{2}}^{6} \frac{x}{2 x - 3} d x

9

解答ステップ

\int_{\frac{3}{2}}^{6} \frac{x}{2 x - 3} d x

u置換積分法を適用する

= \int_{0}^{3} \frac{u ^{2} + 3}{2} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{3} u^{2} + 3 d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{3} u^{2} d u + \int_{0}^{3} 3 d u)

\int_{0}^{3} u^{2} d u = 9

\int_{0}^{3} 3 d u = 9

= \frac{1}{2} (9 + 9)

簡素化

\frac{1}{2} (9 + 9) : 9

= 9